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Magazine Chapter 1: Problem 26
Solve the differential equations in Problem by regarding \(y\) as the independent variable rather than \(x\). $$ \left(1-4 x y^{2}\right) \frac{d y}{d x}=y^{3} $$
Short Answer
Expert verified
\(x = \frac{-\frac{1}{2y^2} + C}{1 + 4\ln y}\), considering \(y\) as the independent variable.
Step by step solution
01
Rearrange to Dy/Dx Form
Start by rearranging the given equation to isolate the differential term \(\frac{dy}{dx}\) on one side: \[\left(1-4xy^2\right) \frac{dy}{dx} = y^3\] which can be rewritten as: \[\frac{dy}{dx} = \frac{y^3}{1-4xy^2}\]
02
Make Substitution by Interchanging Variables
Since we are considering \(y\) as the independent variable, we must express \(x\) in terms of \(y\). This gives us \(\frac{dx}{dy}\) instead of \(\frac{dy}{dx}\). Take the reciprocal of both sides of the equation from Step 1: \[\frac{dx}{dy} = \frac{1-4xy^2}{y^3}\]
03
Separate Variables
Separate the variables \(x\) and \(y\) so that each side of the equation consists of a single variable and its differential: \[dx = \frac{1-4xy^2}{y^3}dy\] Simplify the expression by rewriting it to isolate terms associated with \(x\) and \(y\): \[x = \int \frac{1}{y^3} dy - \int \frac{4x}{y} dy\]
04
Solve the Integrals
Solve the integral \(\int \frac{1}{y^3}dy\) as follows: \[\int \frac{1}{y^3}dy = -\frac{1}{2y^2} + C_1\] where \(C_1\) is a constant of integration. Now solve the term involving \(x\): \[x = -\frac{1}{2y^2} - 4x\int \frac{1}{y}dy + C_1 \]
05
Simplify and Solve for x
Combine the terms and solve for \(x\). Since the equation \(x = -\frac{1}{2y^2} - 4x \ln y + C_1\) is complex, isolate \(x\) terms: \[x + 4x \ln y = -\frac{1}{2y^2} + C_1\] Simplifying further, this becomes \[x(1 + 4\ln y) = -\frac{1}{2y^2} + C_1\] \[x = \frac{-\frac{1}{2y^2} + C_1}{1 + 4\ln y}\]
06
Final Re-arrangement
To find a clearer description of \(x\), factor and adjust constants accordingly: \[x = \frac{-\frac{1}{2y^2} + C_1}{1 + 4\ln y}\] Hence, \(x\) can be expressed in terms of \(y\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Variable Separation
In solving differential equations, variable separation is a critical technique that allows us to simplify complex equations. It involves isolating variables on either side of the equation so that each variable appears only with its differential terms.
To apply this method, rearrange the equation until one side contains only one variable and its corresponding differential. For instance, the equation \(dx = \frac{1 - 4xy^2}{y^3}dy\) can be rewritten by separating terms associated with \(x\) and terms associated with \(y\).
This separation permits integration on both sides of the equation, giving us the pathway to solving the equation completely. Variable separation is effective for differential equations where both variables can be distinctly isolated.
To apply this method, rearrange the equation until one side contains only one variable and its corresponding differential. For instance, the equation \(dx = \frac{1 - 4xy^2}{y^3}dy\) can be rewritten by separating terms associated with \(x\) and terms associated with \(y\).
This separation permits integration on both sides of the equation, giving us the pathway to solving the equation completely. Variable separation is effective for differential equations where both variables can be distinctly isolated.
Independent Variable
In a given differential equation, the independent variable is the variable typically plotted on the horizontal axis or used as the reference in measuring change. For most equations, \(x\) is the independent variable, but this can be interchanged as needed.
In our example, we are treating \(y\) as the independent variable instead of \(x\). This requires applying some clever math techniques, such as expressing \(x\) in terms of \(y\) and taking the reciprocal of the derivative.
Changing the perspective of the independent variable can completely alter our approach to solving differential equations. It requires careful rearrangement and manipulation of the original equation.
In our example, we are treating \(y\) as the independent variable instead of \(x\). This requires applying some clever math techniques, such as expressing \(x\) in terms of \(y\) and taking the reciprocal of the derivative.
Changing the perspective of the independent variable can completely alter our approach to solving differential equations. It requires careful rearrangement and manipulation of the original equation.
Integrals
Integrals play a vital role in solving differential equations, especially after separating variables. They allow us to evaluate areas under curves and to sum small pieces together, providing the solution to the differential equation.
In our exercise, solving the integral \(\int \frac{1}{y^3}dy\) results in \(-\frac{1}{2y^2} + C_1\). This step is critical because it helps us integrate each side of the equation as a function of their respective variables.
Sometimes, the integration might involve logarithmic functions or more complex algebraic manipulations, as evidenced by the second part of the equation in our exercise. Understanding integral calculus is essential for effectively using this method.
In our exercise, solving the integral \(\int \frac{1}{y^3}dy\) results in \(-\frac{1}{2y^2} + C_1\). This step is critical because it helps us integrate each side of the equation as a function of their respective variables.
Sometimes, the integration might involve logarithmic functions or more complex algebraic manipulations, as evidenced by the second part of the equation in our exercise. Understanding integral calculus is essential for effectively using this method.
Constant of Integration
Anytime we perform indefinite integration, we encounter a constant of integration. This constant represents an arbitrary constant that arises from the fact that indefinite integrals represent families of functions.
In our solution, \(C_1\) serves as the constant of integration for \(\int \frac{1}{y^3}dy\). This constant accounts for any vertical shifts in the integral curve, ensuring that the general solution includes all possible specific solutions.
Remembering to include the constant of integration is crucial because it impacts the final solution of the differential equation. Its inclusion ensures the completeness of our mathematics and often plays a role in boundary conditions or initial values in applied problems.
In our solution, \(C_1\) serves as the constant of integration for \(\int \frac{1}{y^3}dy\). This constant accounts for any vertical shifts in the integral curve, ensuring that the general solution includes all possible specific solutions.
Remembering to include the constant of integration is crucial because it impacts the final solution of the differential equation. Its inclusion ensures the completeness of our mathematics and often plays a role in boundary conditions or initial values in applied problems.
